Answer:
he angle between the line BH and the plane AEHD is approximately 56.5 degrees, correct to 1 decimal place.
Explanation:
To find the angle between the line BH and the plane AEHD, we first need to find the vector representing the line BH and the normal vector of the plane AEHD.
The vector representing the line BH is given by the difference between the coordinates of points B and H:
BH = HB = OB - OH = (0, 0, 0) - (8, 23, 23) = (-8, -23, -23)
The normal vector of the plane AEHD can be found by taking the cross product of two vectors in the plane, for example, AE and AD:
AE = EA = EH + HA = (8, 23, 0) + (0, 23, 0) = (8, 46, 0)
AD = (0, 0, 21)
Normal vector of AEHD = AE x AD = (8, 46, 0) x (0, 0, 21) = (-966, 168, 0)
Now, we can find the angle between BH and the plane AEHD using the formula:
cos θ = (BH · n) / (|BH| |n|)
where · represents the dot product, | | represents the magnitude, and n is the normal vector of the plane AEHD.
BH · n = (-8, -23, -23) · (-966, 168, 0) = 19650
|BH| = √((-8)² + (-23)² + (-23)²) = √1338 ≈ 36.6
|n| = √((-966)² + 168² + 0²) = √935460 ≈ 967.5
cos θ = 19650 / (36.6 * 967.5) ≈ 0.540
Therefore, the angle between BH and the plane AEHD is given by:
θ = cos⁻¹(0.540) ≈ 56.5 degrees
So the angle between the line BH and the plane AEHD is approximately 56.5 degrees, correct to 1 decimal place.